TOJ3455 Diamonds

John and Kate found 10 diamonds, each of which has an integer value. They want to divide these diamonds into two parts, one of which will belong to John and the other will belong to Kate. Of cause the two friends both holp to get as more valuable as possible. For equity reason, they will devide these diamonds in such way: at first John divide these diamonds into to part freely, then Kate take one part, and the other part belongs to John. Now the question is how much value can John get at most?

Input

The first line of input is an integer T (1 ≤ T ≤ 100) indicate the number of test cases. Then T lines follows, each line has 10 integers, all of which are integers and between 0 and 1000 (inclusively) represent the value of the ten diamonds.

Output

For each test cases, output a single line: the most value John can get.

Sample Input

11 1 1 2 1 1 1 1 1 1

Sample Output

5

10颗钻石总共有2^9种可能性，直接遍历，这里新学到一个位运算实现遍历

#include <cstdio>#include <iostream>#include <cmath>using namespace std;int main(){    int T;    scanf("%d",&T);    while(T--){        int a[11];        for(int i = 0;i < 10;i++){            scanf("%d",&a[i]);        }        int ans = 0;        for(int i = 0;i < (1 << 9);i++){//这里是精髓，用二进制第i位是0还是1表示a[i]取或者不取            int sum1 = 0,sum2 = 0;            for (int j = 0;j < 10;j++){                if (i & (1 << j)) sum1 += a[j];//如果取a[j]，sum1+=a[i]                    else sum2 += a[j];            }            ans = max(ans,min(sum1,sum2));        }        printf("%d\n",ans);    }    return 0;}